solve this:
1/2 |(bc^2-b^2c)-a(c^2-b^2)+a^2(c-b)|
Answers
Answer:
This question seems to be tricky but it can be solved with a series of cautious steps:
Firstly we need to find out a common factor from the given six terms .
By observation,we found out 2nd & 5th term,and 3rd & 4th term have common factor of (c-a)
Thus our equation becomes:
=>bc^2 -b^2 C+ab^2 -ac^2 +a^2 c-a^2 b
=> bc^2 -b^2 (c-a)-ac(c-a)-a^2 b
Now we can see first and last term can't be in order to solve our problem,we will add and subtract abc at starting and ending respectively in order to take(c-a) common
=>-abc+bc^2 -b^2 (c-a)-ac(c-a)-a^2 b+abc
=>bc(c-a)-b^2 (c-a)-ac(c-a)+ab(c-a)
=>(c-a)(ab-ac+bc-b^2 )
Now take out common from second factor also
=>(c-a)(a(b-c)-b(b-c))
=>(c-a)(b-c)(a-b)
But I would say final answer is
=>(a-b)(b-c)(c-a)
Hope you liked it.
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Let f(a,b,c)=bc2−b2c−ac2+a2c+ab2−a2b=bc(c−b)+ac(a−c)+ab(b−a).
Observe that f vanishes on each of 3 planes a=b,b=c, and a=c in R3 and it's a polynomial of degree 3.
Thus f(a,b,c)=k(a−b)(b−c)(c−a) for some k∈R.
You need only compute k by choosing one point outside of all three planes, for example
(a,b,c)=(1,0,−1). Then f(1,0,−1)=−2k=−2 , so k=1.
Group the first term with the third and factor c^2: (b-a)*c^2
Group the second and the fourth together and factor c: c*(-b^2 + a^2), i.e. c*(a-b)*(a+b)
Group the last two terms together and factor ab: ab*(b-a)
Now when you group these three expressions, you can factor (b-a). Therefore:
(b-a) * (c^2 - c* (a+b) + ab) =
(b-a) * (c^2 - ca - cb + ab) =
(b-a) * (c^2 - cb + ab - ca) =
(b-a)* (c*(c-b) + a*(b-c)) =
(b-a) * (c*(c-b) - a*(c-b)) =
(b-a) * (c-b) *(c-a) =
(a-b) * (b-c) * (c-a)
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How do I prove that ab2+a2b+bc2+b2c+ac2+a2c=(a+b+c)(ab+bc+ac)−3abc ?
How can Cauchy-Schwarz prove that if a, b, c, are positive numbers, then (a2b+b2c+c2a)(ab2+bc2+ca2)≥9a2b2c2 ?
How can you factor this a^2(b-c) +b^2(c-a) +c^2(a-b)?
Consider
bc2−b2c−ac2+a2c+ab2−a2b
=bc2−b2c−ac2+abc−abc+a2c+ab2−a2b
=c(bc−b2−ac+ab)−a(bc−ac−b2+ab)
=(bc−b2−ac+ab)(c−a)
=b(c−b)−a(c−b)(c−a)
=(b−a)(c−b)(c−a)
=(a−b)(b−c)(c−a)
∴bc2−b2c−ac2+a2c+ab2−a2b=(a−b)(b−c)(c−a)
Let ,S=bc2−b2c−ac2+a2c+ab2−a2b
Put a=b , S=0 . Therefore a-b is a factor.
Similarly, b-c and c-a are also the factors.
Therefore, S=k(a-b)(b-c)(c-a). This is an identity which is valid for all numbers.
EDIT: Since the highest degree of the given expression is 3, we do not have any other factor.
Put a=0 ,b=1 and c=2
S=2k=1×2^2−1^2×2−0×2^2+0^2×2+0×1^2−0^2×1=2
So, k=1
Therefore, Given expression =(a-b)(b-c)(c-a)
How do I prove that ab2+a2b+bc2+b2c+ac2+a2c=(a+b+c)(ab+bc+ac)−3abc ?
HELLO, we can prove the evaluation by looking at the RIGHT HAND SIDE of this equation...I have written the answer and I hope it is useful to you :) PEACE
How can Cauchy-Schwarz prove that if a, b, c, are positive numbers, then (a2b+b2c+c2a)(ab2+bc2+ca2)≥9a2b2c2 ?
PS - [Comment on Question v1.] I think you got the wrong inequality Cauchy-Schwarz inequality implies the inequality \text{Arithmetic Mean} \geq \text{Geometric Mean} When we have 3 numbers we get the equation \frac{x+y+z}{3} \geq \left( x y z \right)^{1/3} We then get \frac{a^2 b + b^2 c + c^2 a}{3}
How can you factor this a^2(b-c) +b^2(c-a) +c^2(a-b)?
This is a classic. a^2(b - c) + b^2(c-a) + c^2(a - b) = a^2b - a^2c + b^2c - b^2a + c^2a-c^2b =a^2b - b^2a - a^2c + b^2c + c^2a -c^2b = ab(a - b) - c(a^2 - b^2) +c^2(a - b) =ab(a - b) -c(a - b)(a + b) +c^2(a - b) =(a - b)(ab - c(a + b) +c^2) =(a - b)(ab - cb -ca + c^2) = (a - b)(b(a-c) -c(a-c)) =(a-b)(b-c)(a
How do I factorise 2b2c2+2c2a2+2a2b2−a4−b4−c4 ?
1. (x-y-z)^2=x^2+y^2+z^2-2(xy-yz+xz) 2. x^2-y^2=(x-y)(x+y) 3. (x-y)^2=x^2-2xy+y^2 This identities can be proved easliy by induction method. Hence by using this identities we get 2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4=-[a^4+b^4+c^4-2(a^2b^2-b^2c^2+c^2a^2)]+4b^2c^2 =4b^2c^2-(a^2-b^2-c^2)^2 =(2bc-a^2+b^2+c^2)(2bc+a^2-b^
How can I Factorize 3a2+4ab+b2−2ac−c2 ?
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How do I solve (a+b+c)(a2+b2+c2−ab−bc−ac) ?
I think you mean simplifying so in that perspective there’d probably be two ways to do it, 1 is using foil and multiplying a, b, and c with the other variables, and the simplifying further
How do I prove a3b+b3c+c3a≥a2b2+b2c2+c2a2?
You don't. Because it's false. Take a=1,\,\,b=1,\,\,c=-1 LHS=-1 RHS=3 Hence not true for all a, b, c